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Let $\mathfrak{r}$ be a reductive Lie subalgebra of the complex semisimple Lie algebra $\mathfrak{g}$ and let $B$ be the Killing form of $\mathfrak{g}$.

Suppose that the restriction $B|_\mathfrak{r}$ of B on $\mathfrak{r}$ is non-degenerate. Let $\mathfrak{g} = \mathfrak{r} \oplus\mathfrak{s}$ be the orthogonal decomposition with respect to $B$.

How to check that the restriction $B|_\mathfrak{s}$ of $B$ on $\mathfrak{s}$ is also non-degenerate?

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    $\begingroup$ Hint: the matrix associated to the form is block diagonal when you pick bases from those two subspaces. $\endgroup$ Feb 26, 2019 at 8:52
  • $\begingroup$ The accepted answer is of course correct, but maybe one should notice that this has very little to do with Lie algebras. It's a general fact that if $B$ is a non-degenerate bilinear form on a vector space $V$, and $W$ is a subspace, then the restriction of $B$ to $W$ is non-degenerate if and only if $V = W \oplus W^\perp$. Cf. math.stackexchange.com/q/1295105/96384 or math.stackexchange.com/q/3550381/96384. Here you just use the $\Leftarrow$ direction (which of course is proved most easily as in Tsemo Aristide's answer). $\endgroup$ Mar 16, 2022 at 18:48

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Suppose $g$ is the orthogonal sum $g=r\oplus s$ where $g$ is semi-simple. Since $g$ is sem-simple, Killing is not degenerated, this implies that for every $y\in s$, there exists $z\in g$ such that $B(y,z)\neq 0$, we can write $z=u+v, u\in r, v\in s$, we deduce that $B(y,z)=B(y,u+v)=B(y,v)\neq 0$ since $r$ is orthogonal to $s$. This implies that the restriction of $B$ to $s$ is not degenerated.

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  • $\begingroup$ Thank you for the clear explanation. $\endgroup$ Feb 27, 2019 at 7:50

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